Uniqueness of Sylow 3-Subgroup in a Group of Order 3333

Question

Let $G$ be a finite group of order $|G| = 3333$. How can one prove that $G$ has a unique Sylow $3$-subgroup, i.e., that the Sylow $3$-subgroup is normal in $G$?

I tried to approach this by considering the number of conjugates that the Sylow $3$-subgroup has, but this didn’t lead me anywhere. I also attempted to prove that G must be cyclic, thinking that if I could establish this, the claim would follow trivially—but that approach didn’t work either.

I would appreciate any insights or alternative methods to tackle this problem.

Answer 1

The prime decomposition of $3333$ is $3333 = 3\times 11\times 101$.

Let's start by counting the number $n_{101}$ of Sylow $101$-subgroups. By Sylow's theorems, $n_{101}=1 \pmod{101}$ and $n_{101}|33$.
These two requirements imply $n_{101}=1$ so there is a unique Sylow $101$-subgroup $N$, which is necessarily normal. Now, the group $G$ acts on $N$ via conjugation (since $N$ is normal). Note that this action gives a group homomorphism $\varphi:G\rightarrow\mathrm{Aut}(N)$ where $\mathrm{Aut}(N)$ is the automorphism group of $N$. However, $N$ is isomorphic to the cyclic group $C_{101}$ with $101$ elements and its automorphism group has $100$ elements (see this answer here for a proof). Hence, $|\varphi(G)|$ divides both $3333$ and $100$, which implies that $\varphi(G)$ has only one element, i.e., $G$ acts trivially on $N$. We deduce that $N$ is contained in the center of $G$.

Let $Z(G)$ denote the center of $G$. Then $G/Z(G)$ is a group of cardinality $1,3,11$ or $33$. Hence, $G/Z(G)$ must be cyclic (for $1,3,11$ this is obvious, for $33$ it follows from Sylow's theorems, see this answer). Since $G/Z(G)$ is cyclic, $G$ must be abelian (see this answer). Hence, every Sylow subgroup of $G$ is normal.

Edit: The proof shows that every group of order $3333$ is in fact isomorphic to $C_3\times C_{11}\times C_{101}$. You can find a generalization of this fact in this link, where the answer to the question shows that if $(n,\varphi(n))=1$, then every group of order $n$ is cyclic. This is precisely the same situation as ours ( $(3333,\varphi(3333))=(3333,2000)=1$). The proof of this general fact is analogous to what I wrote here: Start with the largest prime that divides $n$, show that the Sylow subgroup must be unique, then deduce from the order of its automorphism group that it must be central. Iterating the argument you deduce that $G$ is abelian.

Answer 2

It is a fact that if a group has order $n$ and $(n,\varphi(n))=1$ then $G$ is cyclic. Note that $\varphi(3333)=\varphi(3)\varphi(11)\varphi(101)=2\cdot 10\cdot100=2\cdot10^3$. Hence $(3333,\varphi(3333))=1$ and the result follows. Nevertheless, verifying this statement is not straightforward.

A straightforward exercise using Sylow’s theory, however, is to prove that if $|G|=pq$, where $p<q$ are primes and $p$ does not divide $q-1$, then $G$ is cyclic. This is what I'm going to use.

Let $Q\in\text{Syl}_{11}(G)$, $R\in\text{Syl}_{101}(G)$. By Sylow's theory, $|G:\textbf N_G(Q)|\equiv1\text{ (mod }11\text{)}$ and $|G:\textbf N_G(R)|\equiv1\text{ (mod }101\text{)}$ and thus $Q,R\unlhd G$. Then $|G/Q|=3\cdot101$ and $|G/R|=3\cdot11$. As $3$ does not divide either $100$ or $10$, then both $G/Q$ and $G/R$ are cyclic, and in particular they are abelian. As a consequence, the derived subgroup $G'$ is contained in both $Q$ and $R$, and since these subgroups have coprime order then necessarily $G'=1$. This implies that $G$ is abelian, and we are done.

Answer 3

You are rightly suspecting that such a $G$ is abelian. This is actually a particular case of the general setting: $|G|=pqr$, $p<q<r$, $p\nmid q-1$, $p\nmid r-1$ and $q\nmid r-1$. The following argument uses:

1. Sylow III;
2. normal subgroups are union of conjugacy classes;
3. a group of order $st$ ($s,t$ prime numbers), such that $s<t$ and $s\nmid t-1$, is cyclic;
4. $G/H$ cyclic for a central normal $H$ forces $G$ to be abelian.

By Sylow III, $n_p\mid qr$ and $n_p\equiv 1\pmod p$. So, $n_p=1,q,r,qr$ and $n_p=1+kp$. Since $p\nmid q-1$ and $p\nmid r-1$, then $n_p\ne q,r$ and hence $n_p=1,qr$. Likewise, $n_q\mid pr$ and $n_q\equiv 1\pmod q$. So, $n_q=1,p,r,pr$ and $n_q=1+lq$. Now, $q\nmid p-1$ (because $q>p$); so, since $q\nmid r-1$, then $n_q\ne p,r$ and hence $n_q=1,pr$.

Suppose there are $qr$ $p$-Sylow subgroups and $pr$ $q$-Sylow subgroups. Since all these subgroups intesect pairwise trivially (they have order $p$ or $q$), their union's size amounts to $qr(p-1)+pr(q-1)+1$, which exceeds $pqr$$^\dagger$, a contradiction. Therefore, there isn't enough room in $G$ for so many $p$-Sylows and $q$-Sylows at the same time, and hence either $n_p=1$ or $n_q=1$.

If $n_p=1$, then the only $p$-Sylow is normal, say $H$. As such, $H$ is the union of conjugacy classes of $G$, with as many singletons as $|H\cap Z(G)|$. But, by Lagrange, $|H\cap Z(G)|=1,p$, and the former option is ruled out because there aren't conjugacy classes (of sizes $p$, $q$, $r$ and their pairwise products) "filling the gap" $p-1$. Therefore $H$ is central, and being $G/H$ cyclic (as $q\nmid r-1$), then $G$ is abelian.

Same argument in case of $n_q=1$ (say $K$ the only $q$-Sylow, normal), with the only difference that now "filling the gap" $q-1$ is prevented by the assumption $p\nmid q-1$, and the ciclicity of $G/K$ is ensured by $p\nmid r-1$.

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$^\dagger$In fact: $qr(p-1)+pr(q-1)+1=$ $2pqr-r(p+q)+1>pqr\iff$ $pqr-r(p+q)+1>0$, which is true because $pqr-r(p+q)+1>$ $pqr-2qr+1=$ $qr(p-2)+1>0$.